Selective Value and Average Excess Productivity

Under constant organization, molecular selection is a kinetic competition among self-replicating species, not an organism-level fitness contest. Each species i is assigned a selective value W_ii, the net replication rate obtained by subtracting its degradation rate from its replication rate. The population-level average excess productivity E(t) is the composition-weighted mean of these selective values:

E(t) = ∑_j W_jj

Because E(t) depends on the current frequencies or concentrations , selective advantage is relative and dynamic. Species i increases only when its own selective value exceeds the current population average, W_ii > E(t). The selection equation is:

[W_ii − E(t)]

Constant organization is maintained because the weighted excesses sum to zero: ∑_i [W_ii − E(t)] = 0. If W_ii > E(t), species i grows in frequency; if W_ii < E(t), it declines; if W_ii = E(t), it is momentarily neutral. As composition changes, E(t) moves, so the same W_ii can confer advantage early and vanish later. In the simplest selection-only case, the species with the highest W_ii is eventually enriched and E(t) approaches that maximal selective value.

Core Concepts

Selective value W_ii: net replication rate for species i, combining replication and degradation terms. It defines the diagonal selective value used in the selection equations.
Average excess productivity E(t) = ∑_j W_jj : the population average of selective values weighted by species frequencies or concentrations. It changes as composition changes.
Selection equation (W_ii − E(t)): species i increases when W_ii > E(t), declines when W_ii < E(t), and is momentarily neutral when W_ii = E(t).

Two-Species Example

Suppose . Initially E(t) lies between W₂₂ and W₁₁. Species 1 has and rises; species 2 has and falls. As → 1, E(t) → W₁₁, and the net excess of species 1 approaches zero. Selective advantage here is not a fixed property of a molecule; it is the difference between its net replication rate and the current population average.

Common Misconception

A molecule with a positive W_ii does not automatically increase. It must exceed E(t). Likewise, a molecule with a negative W_ii can persist if E(t) is even lower.

Selective Value and Average Excess Productivity

Under constant organization, the species frequencies x 1 and x 2 shift while the average excess productivity E(t) rises from a value between W 22 and W 11 toward W 11, because species 1 comes to dominate. As E(t) climbs, the excess W 11 − E(t) shrinks toward zero, so species 1's growth slows and becomes momentarily neutral as fixation is approached.

QUESTION 1

XVI.2. What does selective advantage mean to a molecule under constant organization?

Its absolute concentration in the reaction volume.

The difference between its net replication rate Wᵢᵢ and the population average excess productivity E(t).

Its cellular fitness relative to the other organisms in the environment.

Its total replication rate, with the degradation term ignored.

QUESTION 2

Which expression defines the average excess productivity E(t) under constant organization?

E(t) = ∑ⱼ Wⱼⱼ xⱼ

, an unweighted mean of the selective values

E(t) = ∑ⱼ xⱼ / Wⱼⱼ

E(t) = ∑ⱼ Wⱼⱼ, the sum of the selective values

QUESTION 3

In the selection equation dxᵢ/dt = xᵢ(Wᵢᵢ − E(t)), why is constant organization maintained?

Because the weighted excesses sum to zero: ∑ᵢ xᵢ(Wᵢᵢ − E(t)) = 0.

Because all species have the same selective value Wᵢᵢ.

Because the frequencies xᵢ are held fixed by the environment.

Because E(t) is fixed by the initial composition and never changes.

QUESTION 4

A molecular species with Wᵢᵢ < E(t) will:

increase in frequency because its replication rate is positive.

stay at constant frequency as long as its Wᵢᵢ does not change.

decline in frequency because its net replication rate is below the population average.

be unaffected, because its own replication rate is unchanged.

QUESTION 5

In a two-species selection-only system with

, what happens to E(t) as

→ 1?

E(t) approaches W₂₂, and species 1 declines.

E(t) remains exactly halfway between W₁₁ and W₂₂.

E(t) approaches W₁₁, and the net excess of species 1 approaches zero.

E(t) increases without bound.

Case Study: Two-Species Prebiotic Reactor

Read the scenario below and answer the questions.

A prebiotic reactor maintains constant total concentration of two self-replicating molecular species A and B. Species A has selective value

min⁻¹ and species B has

min⁻¹. Initially

and

. The selection equation is dxᵢ/dt = xᵢ(Wᵢᵢ − E(t)), with E(t) = ∑ⱼ Wⱼⱼ xⱼ.

1. Calculate the initial average excess productivity E(0). Which species initially increases, and which initially declines?

Answer:
E(0) =

min⁻¹. Species A:

(0) =

, so A increases. Species B:

= −

, so B declines.

2. Explain what happens to E(t) and to the net excess of species A as

→ 1. Why is selective advantage not a fixed property?

Answer:
As

→ 1, E(t) =

+ (

)

approaches

min⁻¹. The net excess

(t) approaches 0, so species A's growth slows and stops at fixation. Selective advantage is not fixed because it depends on composition: the same

confers advantage only while E(t) <

, and the advantage disappears as the population becomes dominated by A. This illustrates relative, dynamic molecular selection rather than absolute fitness.